Question

求sin3xsin5x 原函数，求1/[（sinx)^4+(cosx)^4]的原函数

Answer 1

∫sin3xsin5xdx = (1/2)∫(cos2x-cos8x)dx (用了积化和差）
= (1/2)[(1/2)sin2x-(1/8)sin8x]+C
= sin(2x)/4-sin(8x)/16+C.

(sinx)^4+(cosx)^4 = [(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2
= 1-(1/2)(sin2x)^2 = 1-(1/4)(1-cos4x) = (1/4)(3+cos4x),
则 I =∫dx/[（sinx)^4+(cosx)^4] = ∫d(4x)/(3+cos4x)
半角代换，令 u=tan2x，则 cos4x=(1-u^2)/(1+u^2), d(4x)=2du/(1+u^2)，得
I = ∫du/(2+u^2) =(1/√2)∫d(u/√2)/(1+u^2/2)
= (1/√2)arctan(u/√2)+C = (1/√2)arctan[tan(2x)/√2]+C.